Question

One type of electromagnetic radiation has a frequency of 107.1 MHz, another type has a wavelength of \(2.12 \times 10^{-10} \mathrm{~m}\), and another type of electromagnetic radiation has photons with energy equal to \(3.97 \times 10^{-19} \mathrm{~J} / \mathrm{photon}\). Identify each type of electromagnetic radiation and place them in order of increasing photon energy and increasing frequency.

Short answer

The given electromagnetic radiations are identified and ordered as follows: Type 1 are Radio waves, Type 2 are Gamma rays, and Type 3 are Infrared radiation. Both increasing photon energy and frequency orderings are: Type 1 (Radio waves), Type 3 (Infrared radiation), and Type 2 (Gamma rays).

Step-by-step solution

Calculating Wavelength and Energy (Type 1)

Given frequency \(\nu_1 = 107.1 \, \mathrm{MHz}\) which is \(107.1 \times 10^6 \, \mathrm{Hz}\). We can calculate the wavelength using the speed of light equation: \(\lambda_1 = \frac{c}{\nu_1} = \frac{3.0 \times 10^8 \mathrm{~m/s}}{107.1 \times 10^6 \, \mathrm{Hz}}\) \(\lambda_1 = 2.80 \times 10^{-3} \, \mathrm{m}\) Now, we can calculate the photon energy using the photon energy equation: \(E_1 = h \nu_1 = (6.63 \times 10^{-34} \mathrm{~J \cdot s})(107.1 \times 10^6 \, \mathrm{Hz})\) \(E_1 = 7.09 \times 10^{-27} \, \mathrm{J}\) So, \(E_1 = 7.09 \times 10^{-27} \, \mathrm{J}\) and \(\nu_1 = 107.1 \times 10^6 \, \mathrm{Hz}\)

Calculating Frequency and Energy (Type 2)

Given wavelength \(\lambda_2 = 2.12 \times 10^{-10} \, \mathrm{m}\). We can calculate the frequency using the speed of light equation: \(\nu_2 = \frac{c}{\lambda_2} = \frac{3.0 \times 10^8 \mathrm{~m/s}}{2.12 \times 10^{-10} \, \mathrm{m}}\) \(\nu_2 = 1.42 \times 10^{18} \, \mathrm{Hz}\) Now, we can calculate the photon energy using the photon energy equation: \(E_2 = h \nu_2 = (6.63 \times 10^{-34} \mathrm{~J \cdot s})(1.42 \times 10^{18} \, \mathrm{Hz})\) \(E_2 = 9.42 \times 10^{-16} \, \mathrm{J}\) So, \(E_2 = 9.42 \times 10^{-16} \, \mathrm{J}\) and \(\nu_2 = 1.42 \times 10^{18} \, \mathrm{Hz}\)

Calculating Frequency and Wavelength (Type 3)

Given photon energy \(E_3 = 3.97 \times 10^{-19} \, \mathrm{J}\). We can calculate the frequency using the photon energy equation: \(\nu_3 = \frac{E_3}{h} = \frac{3.97 \times 10^{-19} \mathrm{~J}}{6.63 \times 10^{-34} \mathrm{~J \cdot s}}\) \(\nu_3 = 5.99 \times 10^{14} \, \mathrm{Hz}\) Now, we can calculate the wavelength using the speed of light equation: \(\lambda_3 = \frac{c}{\nu_3} = \frac{3.0 \times 10^8 \mathrm{~m/s}}{5.99 \times 10^{14} \, \mathrm{Hz}}\) \(\lambda_3 = 5.01 \times 10^{-7} \, \mathrm{m}\) So, \(E_3 = 3.97 \times 10^{-19} \, \mathrm{J}\) and \(\nu_3 = 5.99 \times 10^{14} \, \mathrm{Hz}\)

Identifying and Ordering Electromagnetic Radiations

Based on the calculated values, we can now identify and order the electromagnetic radiations: Type 1: Radio waves (with \(E_1 = 7.09 \times 10^{-27} \, \mathrm{J}\) and \(\nu_1 = 107.1 \times 10^6 \, \mathrm{Hz}\)) Type 2: Gamma rays (with \(E_2 = 9.42 \times 10^{-16} \, \mathrm{J}\) and \(\nu_2 = 1.42 \times 10^{18} \, \mathrm{Hz}\)) Type 3: Infrared radiation (with \(E_3 = 3.97 \times 10^{-19} \, \mathrm{J}\) and \(\nu_3 = 5.99 \times 10^{14} \, \mathrm{Hz}\)) Order by increasing photon energy: Type 1, Type 3, Type 2 Order by increasing frequency: Type 1, Type 3, Type 2

Key concepts

Photon Energy

The concept of photon energy is crucial in understanding the behavior of electromagnetic radiation. At a fundamental level, electromagnetic radiation is composed of photons, which are particles of light. Each photon carries a discrete amount of energy that can be calculated using the equation

\( E = h u \)

where \(E\) represents the energy of a single photon, \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \text{J}\cdot\text{s}\)), and \(u\) is the frequency of the electromagnetic radiation. This equation is a cornerstone of quantum physics and illustrates how energy is quantized in electromagnetic waves.

The photon energy dictates a lot about its interactions with matter. Higher energy photons, such as gamma rays, are capable of penetrating through materials and causing ionization, which is the basis for their use in medical treatments and their potential hazard as radiation. In contrast, lower energy photons, like those from radio waves, have non-ionizing radiation which is generally considered less harmful and is used for communication signals.

Frequency and Wavelength Relationship

Understanding the relationship between frequency and wavelength is pivotal in characterizing different types of electromagnetic radiation. The two quantities are inversely related to each other, as described by the equation

\( \lambda = \frac{c}{u} \)

where \(\lambda\) represents the wavelength, \(u\) is the frequency, and \(c\) is the speed of light, which is approximately \(3.0 \times 10^8 \text{m/s}\) in a vacuum. This means that as the frequency of the electromagnetic wave increases, its wavelength decreases, and vice versa.

For instance, radio waves have relatively low frequencies and therefore long wavelengths, making them excellent for broadcasting over long distances without much loss of signal. On the other hand, X-rays have high frequencies and short wavelengths, enabling them to have enough energy to penetrate through soft tissue, but enough to be blocked by denser materials like bones, making them useful in medical imaging.

Speed of Light Equation

The speed of light equation encapsulates one of the most fundamental constants in physics. Regardless of the observer's frame of reference, in a vacuum, light always travels at the same speed, which is roughly \(3.0 \times 10^8 \text{m/s}\). This astonishing fact helps scientists calculate either the frequency or wavelength of electromagnetic waves if one of these properties is known.

Using the speed of light, \(c\), along with the wavelength, \(\lambda\), and the frequency, \(u\), we establish the direct relationship of how these properties interact with each other in the form of the equation:

\( c = \lambda u \)

This equation is foundational to the field of electromagnetics, underpinning technologies such as fibre-optic communication, wireless networks, and microwave ovens. Moreover, by manipulating this equation, we open avenues to understanding how energy travels and interacts with various media, leading to advancements in fields ranging from astronomy to telecommunications.